Chapter 20. Electrochemistry Recommendation: Review Sec. 4.4 (oxidation-reduction reactions) in your textbook 20.1 Oxidation-Reduction Reactions Oxidation-reduction reactions = chemical reactions in which the oxidation state of one or more substance changes (redox reactions). Recall: Oxidation involves loss of electrons (OIL). Reduction involves gain of electrons (RIG). Electrochemistry = branch of chemistry that deals with relationships between electricity and chemical reactions. Consider the spontaneous reaction that occurs when Zn is added to HCl. Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) The oxidation numbers of Zn and H + have changed. The oxidation number of Zn has from 0 to +2. The oxidation number of H has from +1 to 0. Zn is oxidized to Zn 2+ while H + is reduced to H 2. H + causes Zn to be oxidized. H + is the oxidizing agent or oxidant. Examples of good oxidizing agents: H 2 O 2, MnO 4 -, Cr 2 O 7 2-, Ce 4+, O 3, and the halogens. Zn causes H + to be reduced. Zn is the reducing agent or reductant. Examples of good reducing agents: alkali and alkaline earth metals. Note: the reducing agent is oxidized and the oxidizing agent is reduced. Sample Exercise 20.1 (p. 859) The nickel-cadmium (nicad) battery, a rechargeable dry cell used in battery-operated devices, uses the following redox reaction to generate electricity: Cd (s) + NiO 2(s) + 2 H 2 O (l) Cd(OH) 2(s) + Ni(OH) 2(s) Identify the substances that are oxidized and reduced.. - 1 -
Practice Exercise 1 (20.1) What is the reducing agent in the following reaction? i.e. What is oxidized? 2 Br - (aq) + H 2 O 2(aq) + 2 H + (aq) Br 2(aq) + 2 H 2 O (l) a) Br - (aq) b) H 2 O 2(aq) c) H + (aq) d) Br 2(aq) e) Na + (aq) Practice Exercise 2 (20.1) Identify the oxidizing (that which is reduced) and reducing agents (that which is oxidized) in the following oxidation-reduction equation: 2 H 2 O (l) + Al (s) + MnO 4 - (aq) Al(OH) 4 - (aq) + MnO 2(s) 20.2 Balancing Oxidation-Reduction Equations Conservation of mass: The amount of each element present at the beginning of the reaction must be present at the end. Conservation of charge: Electrons are not lost in a chemical reaction. Half-Reactions Half-reactions are a convenient way of separating oxidation and reduction reactions. Sn 2+ (aq) + 2Fe 3+ (aq) Sn 4+ (aq) + 2Fe 2+ (aq) Oxidation half-reaction: Sn 2+ (aq) Sn 4+ (aq) +2e (electrons are a product) Reduction half-reaction: 2Fe 3+ (aq) + 2e 2Fe 2+ (aq) (electrons are a reactant) Practice: Write half-reactions for a reaction between Cu 2+ and Zn metal. Overall: Oxidation: Reduction: - 2 -
Balancing Equations by the Method of Half-Reactions Consider the titration of an acidic solution of Na 2 C 2 O 4 (sodium oxalate, colorless) with KMnO 4 (deep purple). MnO 4 is reduced to Mn 2+ (pale pink) while the C 2 O 4 2 is oxidized to CO 2. The equivalence point is indicated by the presence of a pale pink color. If more KMnO 4 is added, the solution turns purple due to the excess KMnO 4. What is the balanced chemical equation for this reaction? We can determine this using the method of half-reactions: 1. Write the overall unbalanced reaction (if available). 2. (i) Identify oxidized and reduced substances (ii) Write down the two incomplete half reactions. MnO 4 (aq) Mn 2+ (aq) C 2 O 4 2 (aq) CO 2 (g) 3. Balance each half reaction: (a) First, balance elements other than H and O. MnO 4 (aq) Mn 2+ (aq) C 2 O 4 2 (aq) 2CO 2 (g) (b) Then balance O atoms by adding water on side deficient in O atoms. MnO 4 (aq) Mn 2+ (aq) + 4H 2 O(l) C 2 O 4 2 (aq) 2CO 2 (g) (c) Then balance H by adding H + for acidic solutions (or OH - for alkaline solutions). 8H + (aq) + MnO 4 (aq) Mn 2+ (aq) + 4H 2 O(l) C 2 O 4 2 (aq) 2CO 2 (g) (d) Finish by balancing charge by adding electrons. This is an easy place to make an error! For the permanganate half-reaction, note that there is a charge of 7+ on the left and 2+ on the right. 5 electrons need to be added to the left: 5e + 8H + (aq) + MnO 4 (aq) Mn 2+ (aq) + 4H 2 O(l) In the oxalate half-reaction, there is a 2 charge on the left and a 0 charge on the right, so we need to add two electrons to the products: C 2 O 4 2 (aq) 2CO 2 (g) + 2e - 3 -
4. Multiply each half-reaction by the appropriate factors so that the number of electrons gained equals electrons lost. To balance the 5 electrons for permanganate and 2 electrons for oxalate, we need 10 electrons for both. Multiplying gives: 10e + 16H + (aq) + 2MnO 4 (aq) 2Mn 2+ (aq) + 8H 2 O(l) 5C 2 O 4 2 (aq) 10CO 2 (g) + 10e 5. Now add the reactions and simplify. 16H + (aq) + 2MnO 4 (aq) + 5C 2 O 4 2 (aq) 2Mn 2+ (aq) + 8H 2 O(l) + 10CO 2 (g) The equation is now balanced! Note that all of the electrons have cancelled out! Sample Exercise 20.2 (p. 862) Complete and balance the following equation by the method of half-reactions: Cr 2 O 7 2- (aq) + Cl - (aq) Cr 3+ (aq) + Cl 2(g) (acidic solution) - 4 -
Practice Exercise 1 (20.2) If you complete and balance the following equation in acidic solution Mn 2+ (aq) + NaBiO 3(s) Bi 3+ (aq) + MnO 4 - (aq) + Na + (aq) how many water molecules are there in the balanced equation (for the reaction balanced with the smallest whole-number coefficients)? a) Four on the reactants side b) Three on the product side c) One on the reactant side d) Seven on the product side e) Two on the product side Practice Exercise 2 (20.2) Complete and balance the following equation in acidic solution using the method of half-reactions. Cu (s) + NO 3 - (aq) Cu 2+ (aq) + NO 2(g) Balancing Equations for Reactions Occurring in Basic Solution The same method as above is used, but OH is added to neutralize the H + used. The equation must again be simplified by canceling like terms on both sides of the equation. - 5 -
Sample Exercise 20.3 (p. 864) Complete and balance this equation for a redox reaction that takes place in basic solution: CN - (aq) + MnO 4 - (aq) CNO - (aq) + MnO 2(s) (basic solution) Practice Exercise 1 (20.3) If you complete and balance the following oxidation-reduction reaction in basic solution NO 2 - (aq) + Al (s) NH 3(aq) + Al(OH) 4 - (aq) how many hydroxide ions are there in the balanced equation (for the reaction balanced with the smallest wholenumber coefficients)? a) One on the reactant side b) One on the product side c) Four on the reactant side d) Seven on the product side e) none Practice Exercise 2 (20.3) Complete and balance the following oxidation-reduction reaction in basic solution: Cr(OH) 3(s) + ClO - (aq) CrO 4 2- (aq) + Cl 2(g) - 6 -
20.3 Voltaic (Galvanic) Cells generates electrical energy from a spontaneous redox reaction can be used to do work IF the half reactions are kept separate electron transfer occurs via an external circuit must provide a conductor for transfer of electrons plus a salt bridge to keep the separate half-reactions electrically neutral anode is where oxidation takes place (electrons are released, so it is the negative electrode) electrons move from the anode to the cathode (where reduction takes place electrons are attracted to the positive electrode) If a strip of Zn is placed in a solution of CuSO 4, Cu is deposited on the Zn and the Zn dissolves by forming Zn 2+. Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Zn is spontaneously oxidized to Zn 2+ by Cu 2+. The Cu 2+ is spontaneously reduced to Cu 0 by Zn. The entire process is spontaneous. - 7 -
Figure 20.5 Voltaic Cell e.g. CuNO 3(aq) with a Cu electrode (cathode) and ZnNO 3(aq) with a Zn electrode (anode) This voltaic cells consists of: An oxidation half-reaction: Zn(s) Zn 2+ (aq) + 2e Oxidation takes place at the anode. A reduction half-reaction: Cu 2+ (aq) + 2e Cu(s) Reduction takes place at the cathode. The two solid metals are the electrodes (cathode and anode). As oxidation occurs, Zn is converted to Zn 2+ and 2e. The electrons flow towards the cathode where they are used in the reduction reaction. We expect the Zn electrode to lose mass and the Cu electrode to gain mass. Electrons flow from the anode to the cathode. the anode is negative and the cathode is positive. Electrons cannot flow through the solution; they have to be transported through an external wire. Anions and cations move through a porous barrier or salt bridge. A salt bridge is filled with a saturated solution of an electrolyte and is used to maintain the electrical neutrality of the solutions in both half-cells. Cations from the salt bridge electrolyte move towards the cathode to neutralize the excess of negatively charged ions (Cathode: Cu 2+ + 2e Cu, so the counter ion of Cu, e.g., NO 3, is in excess). Anions from the salt bridge electrolyte move towards the anode to neutralize the excess Zn 2+ ions formed by oxidation. - 8 -
A summary of the terminology used to describe voltaic cells. Oxidation occurs at the anode; reduction occurs at the cathode. The electrons flow spontaneously from the negative anode to the positive cathode. The movement of ions in solution completes the electrical circuit. Anions move toward the anode, whereas cations move toward the cathode. The cell compartments can be separated by either a porous glass barrier or by a salt bridge (as in Figure on p. 8). Sample Exercise 20.4 (p. 867) The following oxidation-reduction reaction is spontaneous: Cr 2 O 7 2- (aq) + 14 H + (aq) + 6 I - (aq) 2 Cr 3+ (aq) + 3 I 2(s) + 7 H 2 O (l) A solution containing K 2 Cr 2 O 7 and H 2 SO 4 is poured into one beaker, and a solution of KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution (such as platinum foil) is suspended in each solution, and the two conductors are connected with wires through a voltmeter or some other device to detect an electric current. The resultant voltaic cell generates an electric current. Indicate the reaction occurring at the anode, the reaction at the cathode, the direction of electron and ion migrations, and the signs of the electrodes. - 9 -
The following two half-reactions occur in a voltaic cell: Practice Exercise 1 (20.4) Ni (s) Ni 2+ (aq) + 2 e - Cu 2+ (aq) + 2 e - Cu (s) (electrode = Ni) (electrode = Cu) Which one of the following descriptions most accurately describes what is occurring in the half-cell containing the Cu electrode and Cu 2+ (aq) solution? a) The electrode is losing mass and cations from the salt bridge are flowing into the half-cell. b) The electrode is is gaining mass and cations from the salt bridge are flowing into the half-cell. c) The electrode is losing mass and anions from the salt bridge are flowing into the half-cell. d) The electrode is is gaining mass and anions from the salt bridge are flowing into the half-cell. The two half-reactions in a voltaic cell are Zn (s) Zn 2+ (aq) + 2 e - ClO 3 - (aq) + 6 H + (aq) + 6 e - Cl - (aq) + 3 H 2 O (l) Practice Exercise 2 (20.4) a) Indicate which reaction occurs at the anode and which at the cathode. b) Which electrode is consumed in the cell reaction? c) Which electrode is positive? - 10 -
A Molecular View of the Electrode Process Rules of voltaic cells: At the anode electrons are products. Oxidation occurs at the anode. Any ions that are getting oxidized (usually, but not always, anions) flow towards the anode. Any ions that are products of oxidation flow away from the anode. At the cathode electrons are reagents. Reduction occurs at the cathode. Any ions that are getting reduced (usually, but not always, cations) flow towards the anode. Any ions that are products of reduction flow away from the cathode. The flow of electrons from anode to cathode requires an external wire. The transfer of ions through a salt bridge maintains overall charge balance for the two compartments. - 11 -
Figure 20.8 On the left hand side, the anode, the Zn is oxidized, releasing 2 electrons which migrate through the wire to the cathode (Cu electrode). At the cathode the Cu 2+ in solution is reduced and deposited onto the Cu electrode. The counter anion moves from the Cu compartment to the Zn compartment to balance the newlyproduced Zn 2+ ions in the Zn compartment. Zn 2+ ions move into the Cu compartment to balance the excess anions leftover after the Cu 2+ is reduced. 20.4 Cell EMF The flow of electrons from anode to cathode is spontaneous, but what is the driving force? Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than the anode. Potential difference = difference in electrical potential. The potential difference is measured in volts. One volt (V) = the potential difference required to impart one joule (J) of energy to a charge of one coulomb (C): J 1 V = 1 C Electromotive force (emf) = the force required to push electrons through the external circuit. Cell potential: E cell = emf of a cell. = the cell voltage. E cell is > 0 for a spontaneous reaction. For 1M solutions or 1 atm pressure for gases at 25 o C (standard conditions), the standard emf (standard cell potential) is called E o cell. e.g. for the reaction: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) E o cell = +1.10 V - 12 -
Standard Reduction (Half-Cell) Potentials We can conveniently tabulate electrochemical data. Standard reduction potentials, E o red are measured relative to a standard. The emf of a cell can be calculated from standard reduction potentials: We use the following half-reaction as our standard: E o cell= E o red(cathode) E o red (anode) 2H + (aq, 1M) + 2e H 2 (g, 1 atm) E o cell = 0V. This electrode is called a standard hydrogen electrode (SHE) or the normal hydrogen electrode (NHE). The SHE is assigned a standard reduction potential of zero. Figure 20.9-13 -
Example: Consider the half-reaction: Zn(s) Zn 2+ (aq) + 2e We can measure E o cell relative to the SHE (cathode): = a Pt electrode in a tube placed in 1 M H + solution. H 2 is bubbled through the tube. E o cell = E o red(cathode) E o red (anode) 0.76 V = 0 V E o red (anode) E o red (anode) = 0.76 V. Standard reduction potentials must be written as reduction reactions: Zn 2+ (aq, 1M) + 2e Zn(s) E o red = 0.76 V. Since E o red = 0.76 V we conclude that the reduction of Zn 2+ in the presence of the SHE is not spontaneous. However, the oxidation of Zn with the SHE is spontaneous. - 14 -
The standard reduction potential is an intensive property. Changing the stoichiometric coefficient does not affect E o red. 2Zn 2+ (aq) + 4e 2Zn(s) E o red = 0.76 V Reactions with E o red > 0 are spontaneous reductions relative to the SHE. Reactions with E o red < 0 are spontaneous oxidations relative to the SHE. The larger the difference between E o red values, the larger E o cell. The more positive the value of E o red, the greater the driving force for reduction. Sample Exercise 20.5 (p. 872) For the Zn-Cu 2+ voltaic cell shown in Figure 20.5, we have Zn (s) + Cu 2+ (aq, 1M) Zn 2+ (aq, 1M) + Cu (s) E o cell = 1.10 V Given that the standard reduction potential of Zn 2+ to Zn is -0.76 V, calculate the E o red for the reduction of Cu 2+ to Cu. Cu 2+ (aq, 1M) + 2 e - Cu (s) (0.34 V) Practice Exercise 1 (20.5) A voltaic cell based on the reaction 2 Eu 2+ (aq) + Ni 2+ (aq) 2 Eu 3+ (aq) + Ni (s) generates E o = 0.07 V. Given the standard reduction potential of Ni 2+ given in Table 20.1 (-0.28 V) what is he standard reduction potential for the reaction Eu 3+ (aq) + e - Eu 2+ (aq)? a) -0.35 V b) 0.35 V c) -0.21 V d) 0.21 V e) 0.07 V A voltaic cell is based on the following half-reactions: In + (aq) In 3+ (aq) + 2 e - Br 2(l) + 2 e - 2 Br - (aq) Practice Exercise 2 (20.5) The standard emf for this cell is 1.46 V. Using the data in Table 20.1, calculate E o red for the reduction of In 3+ to In +. (-0.40 V) - 15 -
Sample Exercise 20.6 (p. 872) Using the standard reduction potentials listed in Table 20.1, calculate the standard emf for the voltaic cell described in Sample Exercise 20.4, which is based on the following reaction: Cr 2 O 7 2- (aq) + 14 H + (aq) + 6 I - (aq) 2 Cr 3+ (aq) + 3 I 2(s) + 7 H 2 O (l) (0.79 V) Practice Exercise 1 (20.6) Using the data in Table 20.1 what value would you calculate for the standard emf (E o cell) for a voltaic cell that employs the overall cell reaction 2 Ag + (aq) + Ni (s) 2 Ag (s) + Ni 2+ (aq)? a) +0.52 V b) -0.52 V c) +1.08 V d) -1.08 V e) +0.80 V Practice Exercise 2 (20.6) Using the data in Table 20.1, calculate the standard emf for a cell that employs the following overall cell reaction: (+2.20 V) 2 Al (s) + 3 I 2(s) 2 Al 3+ (aq) + 6 I - (aq) - 16 -
Standard cell potential of a voltaic cell. The cell potential measures the difference in the standard reduction potentials of the cathode and the anode reactions: E cell = E red (cathode) E red (anode). In a voltaic cell the cathode reaction is always the one that has the more positive (or less negative) value for E red. Sample Exercise 20.7 (p. 873) A voltaic cell is based on the following two standard half-reactions: Cd 2+ (aq) + 2 e - Sn 2+ (aq) + 2 e - Cd (s) Sn (s) By using the data in Appendix E, determine a) the half-reactions that occur at the cathode and the anode, and b) the standard cell potential (0.267 V) - 17 -
Practice Exercise 1 (20.7) Consider three voltaic cells, each similar to the one shown in Figure 20.5. In each voltaic cell, one half-cell contains a 1.0 M Fe(NO 3 ) 2(aq) solution with an Fe electrode. The contents of the other half-cells are as follows: Cell 1: a 1.0 M CuCl 2(aq ) solution with a Cu electrode Cell 2: a 1.0 M NiCl 2(aq) solution with a Ni electrode Cell 3: a 1.0 M ZnCl 2(aq) solution with a Zn electrode In which voltaic cell(s) does iron act as the anode? a) Cell 1 b) Cell 2 c) Cell 3 d) Cells 1 and 2 e) All three cells. Practice Exercise 2 (20.7) A voltaic cell is based on a Co 2+ /Co half-cell and an AgCl/Ag half-cell. a) Which half-reaction occurs at the anode? b) What is the standard cell potential? (0.499 V) - 18 -
Strengths of Oxidizing and Reducing Agents Consider a table of standard reduction potentials. We can use this table to determine the relative strengths of reducing (and oxidizing) agents. The more positive the E o red the stronger the oxidizing agent (written in the table as a reactant) The more negative the E o red the stronger the reducing agent (written as a product in the table). We can use this to predict if one reactant can spontaneously oxidize another. e.g. F 2 can oxidize H 2 or Li. Ni 2+ can oxidize Al(s). We can use this table to predict if one reactant can spontaneously reduce another. e.g. Li can reduce F 2. - 19 -
Sample Exercise 20.8 (p. 875) Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents: NO 3 - (aq), Ag + (aq), Cr 2 O 7 2- (aq). Practice Exercise 1 (20.8) Based on the data in Table 20.1, which of the following species would you expect to be the strongest oxidizing agent (i.e. most likely to be reduced)? a) Cl - (aq) b) Cl 2(g) c) O 2(g ) d) H + (aq) e) Na + (aq) Practice Exercise 2 (20.8) Using Table 20.1, rank the following species from the strongest to the weakest reducing agent (i.e. most likely to be reduced to the least likely to be reduced): I - (aq), Fe (s), Al (s). 20.5 Free Energy and Redox Reactions For any electrochemical process E o = E o red(reduction process) E o red(oxidation process). A positive E o indicates a spontaneous process (galvanic cell). A negative E o indicates a nonspontaneous process. The above equation is used to understand the activity series of metals. Consider the reaction of nickel with silver ion: Ni(s) + 2Ag + (aq) Ni 2+ (aq) + 2Ag(s) The standard cell potential is: E o cell = E o red(ag + /Ag) E o red(ni 2+ /Ni) = (0.80 V) ( 0.28 V) = 1.08 V This value indicates that the reaction is spontaneous. - 20 -
Sample Exercise 20.9 (p. 876) Using standard reduction potentials (Table 20.1), determine whether the following reactions are spontaneous under standard conditions: a) Cu (s) + 2 H + (aq) Cu 2+ (aq) + H 2(g) b) Cl 2(g) + 2 I - (aq) 2 Cl - (aq) + I 2(s) Practice Exercise 1 (20.9) Which of the following elements is capable of oxidizing Fe 2+ (aq) ions to Fe 3+ (aq) ions: chlorine, bromine, iodine? a) I 2 b) Cl 2 c) Cl 2 and I 2 d) Cl 2 and Br 2 e) all three elements Practice Exercise 2 (20.9) Using standard reduction potentials (Appendix E), determine whether the following reactions are spontaneous under standard conditions: a) I 2(s) + 5 Cu 2+ (aq) + 6 H 2 O (l) 2 IO 3 - (aq) + 5 Cu (s) + H + (aq) b) Hg 2+ (aq) + 2 I - (aq) Hg (l) + I 2(s) c) H 2 SO 3(aq) + 2 Mn (s) + 4 H + (aq) S (s) + 2 Mn 2+ (aq) + 3 H 2 O (l) EMF and G We can show that: G = - nfe Where G is the change in free energy, n is the number of moles of electrons transferred, F is Faraday's constant and E is the emf of the cell. We define a faraday (F) as: C J 1F = 96,500 = 96,500 - - mol e ( V)( mol e ) Since n and F are positive, if G < 0 then E > 0, and the reaction will be spontaneous. When the reactants and products are in their standard states: G o = nfe o - 21 -
Sample Exercise 20.10 (p. 878) a) Use the standard reduction potentials listed in Table 20.1 to calculate the standard free-energy change, G o, and the equilibrium constant, K, at 298 K for the following reaction: 4 Ag (s) + O 2(g) + 4 H + (aq) 4 Ag + (aq) + 2 H 2 O (l) ( G o = -170 kj/mol, K = 9 x 10 29 ) b) Suppose the reaction in part (a) were written as 2 Ag (s) + ½ O 2(g) + 2 H + (aq) 2 Ag + (aq) + H 2 O (l) What are the values of E o, G o and K when the reaction is written in this way? (E o = +0.43 V, G o = - 83 kj/mol, K = 4 x 10 14 ) Practice Exercise 1 (20.10) For the reaction 3 Ni 2+ (aq) + 2 Cr(OH) 3(s) + 10 OH - (aq) 3 Ni (s) + 2 CrO 4 2- (aq) + 8 H 2 O (l) G o = +87 kj/mol. Given the standard reduction potential of Ni 2+ (aq) in Table 20.1, what value do you calculate for the standard reduction potential of the half-reaction CrO 4 2- (aq) + 4 H 2 O (l) + 3 e - Cr(OH) 3(s) + 5 OH - (aq)? a) -0.43 V b) -0.28 V c) 0.02 V d) -0.13 V e) -0.15 V Practice Exercise 2 (20.10) Consider the reaction 2 Ag + (aq) + H 2(g) 2 Ag (s) + 2 H + (aq) Calculate G o f for the Ag + (aq) ion from the standard reduction potentials in Table 20.1 and the fact that G o f for H 2(g), Ag (s) and H + (aq) are all zero. Compare your answer with the value given in Appendix C. - 22 -
20.6 Cell EMF Under Nonstandard Conditions A voltaic cell is functional until E = 0 at which point equilibrium has been reached. The cell is then dead. The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction. The Nernst Equation We can calculate the cell potential under nonstandard conditions. Recall that: G = G o + RTlnQ We can substitute in our expression for the free energy change: nfe = nfe o + RTlnQ Rearranging, we get the Nernst equation: RT E = E ln Q nf or 2.303RT E = E logq nf Note the change from natural logarithm to log base 10. The Nernst equation can be simplified by collecting all the constants together and using a temperature of 298 K: 0.0592 E = E logq n Example: If we have the reaction: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) If [Cu 2+ ] = 5.0 M and [Zn 2+ ] = 0.050M: 0.0592 0.050 = 1.10V log = 1. 2 5.0 Ecell 16 V Sample Exercise 20.11 (p. 881) Calculate the emf at 298 K generated by a voltaic cell in which the reaction is Cr 2 O 7 2- (aq) + 14 H + (aq) + 6 I - (aq) 2 Cr 3+ (aq) + 3 I 2(s) + 7 H 2 O (l) when [Cr 2 O 7 2- ] = 2.0 M, [H + ] = 1.0 M, [I - ] = 1.0 M, and [Cr 3+ ] = 1.0 x 10-5 M. (E = +0.89 V) - 23 -
Practice Exercise 1 (20.11) Consider a voltaic cell whose overall reaction is Pb 2+ (aq) + Zn (s) Pb (s) + Zn 2+ (aq). What is the emf generated by this voltaic cell when the ion concentrations are [Pb 2+ ] = 1.5 x 10-3 M and [Zn 2+ ] = 0.55 M? a) 0.71 V b) 0.56 V c) 0.49 V d) 0.79 V e) 0.64 V Practice Exercise 2 (20.11) For the Zn-Cu voltaic cell depicted in Figure 20.5 would the emf increase, decrease, or stay the same, if you increased the Cu 2+ (aq) concentration by adding CuSO 4. 5H 2 O to the cathode compartment? - 24 -
Sample Exercise 20.12 (p. 882) If the voltage of a Zn-H + cell (like that in Figure 20.9) is 0.45 V at 25 o C when [Zn 2+ ] = 1.0 M and P H2 = 1.0 atm, what is the ph of the cathode solution? (5.2) Practice Exercise 1 (20.12) Consider a voltaic cell where the anode half-reaction is Zn (s) Zn 2+ (aq) + 2 e - and the cathode half-reaction is Sn 2+ (aq) + 2 e - Sn (s). What is the concentration of Sn 2+ if Zn 2+ is 2.5 x 10-3 M and the cell emf is 0.660 V? Use the reduction potentials in Appendix E that are reported to three significant figures. a) 3.3. x 10-2 M b) 1.9 x 10-4 M c) 9.0 x 10-3 M d) 6.9 x 10-4 M e) 7.6 x 10-3 M Practice Exercise 2 (20.12) What is the ph of the solution in the cathode compartment of the cell pictured in Figure 20.9 when P H2 = 1.0 atm, [Zn 2+ ] in the anode compartment is 0.10 M, and cell emf is 0.542 V? (ph = 4.23) - 25 -
Concentration Cells A concentration cell is one whose emf is generated solely because of a concentration difference. Example: Consider a cell with two compartments, each with a Ni(s) electrode but with difference concentrations of Ni 2+ (aq). One cell has [Ni 2+ ] = 1.0 M and the other has [Ni 2+ ] = 0.001 M. The standard cell potential is zero. But this cell is operating under nonstandard conditions! The driving force is the difference in Ni 2+ concentrations. Anode (dilute Ni 2+ ): Ni(s) Ni 2+ (aq) + 2e Cathode (concentrated Ni 2+ ): Ni 2+ (aq) + 2e Ni(s) Using the Nernst equation we can calculate a cell potential of +0.0888 V for this concentration cell. Concentration cell based on the Ni 2+ -Ni cell reaction. In (a) the concentrations of Ni 2+ (aq) in the two compartments are unequal, and the cell generates an electrical current. The cell operates until the concentrations of Ni 2+ (aq) in the two compartments become equal, (b) at which point the cell has reached equilibrium and is "dead". - 26 -
Sample Exercise 20.13 (p. 885) A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has P H2 = 1.00 atm and an unknown concentration of H + (aq). Electrode 2 is a standard hydrogen electrode ([H + ] = 1.00 M, P H2 = 1.00 atm). At 298 K the measured cell voltage is 0.211 V, and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. Calculate [H + ] for the solution at electrode 1. What is its ph? (ph = 3.57) Practice Exercise 1 (20.13) A concentration cell constructed from two hydrogen electrodes, both with P H2 = 1.00 M. One electrode is immersed in pure H 2 O and the other in 6.0 M hydrochloric acid. What is the emf generated by the cell and what is the identity of the electrode that is immersed in hydrochloric acid? a) -0.23 V, cathode b) 0.46 V, anode c) 0.023 V, anode d) 0.23 V, cathode e) 0.23 V, anode Practice Exercise 2 (20.13) A concentration cell is constructed with two Zn (s) -Zn 2+ (aq) half-cells. The first half-cell has [Zn 2+ ] = 1.35 M, and the second half-cell has [Zn 2+ ] = 3.75 x 10-4 M. a) Which half-cell is the anode of the cell? b) What is the emf of the cell? (0.105 V) - 27 -
Cell EMF and Chemical Equilibrium A system is at equilibrium when G = 0. From the Nernst equation, at equilibrium and 298 K (E = 0 V and Q = K eq ) 0.0592 0 = E log Keq n ne log Keq = 0.0592 Thus, if we know the cell emf, we can calculate the equilibrium constant. 20.9 Electrolysis Electrolysis reactions are nonspontaneous reactions that require an external current in order to force the reaction to proceed. They take place in electrolytic cells. In voltaic and electrolytic cells, reduction occurs at the cathode, and oxidation occurs at the anode. However, in electrolytic cells, electrons are forced to flow from the anode to the cathode. In electrolytic cells the anode is positive and the cathode is negative. In voltaic cells the anode is negative and the cathode is positive. Example, decomposition of molten NaCl. Cathode: 2Na + (l) + 2e 2Na(l) (reduction) Anode: 2Cl (l) Cl 2 (g) + 2e. (oxidation) Industrially, electrolysis is used to produce metals like Al. Electrolysis of high-melting ionic substances requires very high temperatures. - 28 -
Do we get the same products if we electrolyze an aqueous solution of the salt? Water complicates the issue! Example: Consider the electrolysis of NaF(aq): Na + (aq) + e Na(s) E o red = 2.71 V 2H 2 O(l) + 2e H 2 (g) + 2OH (aq) Thus water is more easily reduced the sodium ion. E o red = 0.83 V 2F (aq) F 2 (g) + 2e E o red = +2.87 V 2H 2 O(l) O 2 (g) + 4H + (aq) + 4e Thus it is easier to oxidize water than the fluoride ion. E o red = +1.23 V Electrolysis with Active Electrodes Active electrodes: electrodes that take part in electrolysis. Example: electroplating. Consider an active Ni electrode and another metallic electrode (steel) placed in an aqueous solution of NiSO 4 : Anode (nickel strip): Ni(s) Ni 2+ (aq) + 2e Cathode (steel strip): Ni 2+ (aq) + 2e Ni(s). Ni plates on the inert electrode. Electroplating is important in protecting objects from corrosion. Electrolytic cell with an active metal electrode. Nickel dissolves from the anode to form Ni 2+ (aq). At the cathode Ni 2+ (aq) is reduced and forms a nickel plate on the cathode. Quantitative Aspects of Electrolysis We want to know how much material we obtain with electrolysis. Consider the reduction of Cu 2+ to Cu. Cu 2+ (aq) + 2e Cu(s). Two mol of electrons will plate 1 mol of Cu. The charge of one mol of electrons is 96,500 C (1 F). A coulomb is the amount of charge passing a point in one second when the current is one ampere. The amount of Cu can be calculated from the current (I) and time (t) required to plate. Q = I t - 29 -
Sample Exercise 20.14 (p. 895) Calculate the number of grams of aluminum produced in 1.00 hr by the electrolysis of molten AlCl 3 if the electrical current is 10.0 A. (3.3.6 g Al) Practice Exercise 1 (20.14) How much time is needed to deposit 1.0 g of chromium metal from an aqueous solution of CrCl 3 using a current of 1.5 A? a) 3.8 x 10-2 s b) 21 min. c) 62 min. d) 139 min. e) 3.2 x 10 3 min. Practice Exercise 2 (20.14) a) The half-reaction for formation of magnesium metal upon electrolysis of molten MgCl 2 is Mg 2+ + 2e - Mg. Calculate the mass of magnesium formed upon passage of a current of 60.0 A for a period of 4.00 x 10 3 s. (30.2 g Mg) b) How many seconds would be required to produce 50.0 g of Mg from MgCl 2 if the current is 100.0 A? (3.97 x 10 3 s) - 30 -
Electrochemistry Terminology Current = a flow of electrical charge through a conductor. The current at a particular cross-section is the rate of flow of the charge. The charge may be carried by electrons (in metals), ions (electrolyte solutions) or positive holes (semi-conductors). Measured in amperes. Coulomb is the central unit for electrochemistry, enables us to maneuver amongst charge, voltage, current, watts, electrons and mass. The Faraday relates the charge to the number of electrons. 1 V = 1 J/C 1 A = 1 C/s 1 W = 1 J/s 1 F = 96,500 C/mol e - Electrical work: energy units x time 1 watt = 1 J/s 1 watt = 1 V x 1 A # Watts = volts x amps or J x C C s The K sp at 298 K for iron(ii) fluoride is 2.4 x 10-6. Sample Integrative Exercise 20 (p. 897) a) Write a half-reaction that gives the likely products of the two-electron reduction of FeF 2(s) in water. b) Use the K sp value and the standard reduction potential of Fe 2+ (aq) to calculate the standard reduction potential for the half-reaction in part (a). (-0.606 V) c) Rationalize the difference in the reduction potential for the half-reaction in part (a) with that for Fe 2+ (aq). - 31 -